Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $p = \dfrac{-3k^2 + 27}{k^3 - 11k^2 + 24k} \times \dfrac{2k^2 - 16k}{-4k - 40} $
Solution: First factor out any common factors. $p = \dfrac{-3(k^2 - 9)}{k(k^2 - 11k + 24)} \times \dfrac{2k(k - 8)}{-4(k + 10)} $ Then factor the quadratic expressions. $p = \dfrac {-3(k - 3)(k + 3)} {k(k - 3)(k - 8)} \times \dfrac {2k(k - 8)} {-4(k + 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { -3(k - 3)(k + 3) \times 2k(k - 8)} { k(k - 3)(k - 8) \times -4(k + 10)} $ $p = \dfrac {-6k(k - 3)(k + 3)(k - 8)} {-4k(k - 3)(k - 8)(k + 10)} $ Notice that $(k - 3)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-6k\cancel{(k - 3)}(k + 3)(k - 8)} {-4k\cancel{(k - 3)}(k - 8)(k + 10)} $ We are dividing by $k - 3$ , so $k - 3 \neq 0$ Therefore, $k \neq 3$ $p = \dfrac {-6k\cancel{(k - 3)}(k + 3)\cancel{(k - 8)}} {-4k\cancel{(k - 3)}\cancel{(k - 8)}(k + 10)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $p = \dfrac {-6k(k + 3)} {-4k(k + 10)} $ $ p = \dfrac{3(k + 3)}{2(k + 10)}; k \neq 3; k \neq 8 $